Question: Each of the numbers $a_1,$ $a_2,$ $\dots,$ $a_{95}$ is $\pm 1.$  Find the smallest possible positive value of
\[\sum_{1 \le i < j \le 95} a_i a_j.\]
Answer: Let $m$ and $n$ denote the number of 1's and $-1$'s among the $a_i,$ respectively.  Then $m + n = 95$ and
\[a_1^2 + a_2^2 + \dots + a_{95}^2 = 95.\]Let
\[S = \sum_{1 \le i < j \le 95} a_i a_j.\]Then
\[2S + 95 = (a_1 + a_2 + \dots + a_{95})^2 = (m - n)^2.\]Note that $m - n = m + n - 2n = 95 - 2n$ is odd, so $(m - n)^2$ is an odd perfect square.  To minimize $S,$ while still keeping it positive, we take $(m - n)^2$ as the smallest odd perfect square greater than 95, which is 121.  Then $S = \frac{121 - 95}{2} = 13.$

Equality occurs when $m = 53$ and $n = 42,$ so the smallest possible positive value of $S$ is $\boxed{13}.$